This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A352408 #12 Mar 26 2022 15:18:48
%S A352408 0,3,105,4521,247509,16743276,1358620365,129050741220,14072813590533,
%T A352408 1734613925513373,238643675398832787,36267490290699148842,
%U A352408 6035968171216586764461,1092097033048311239037276,213473299061006718358241931,44838030511923231603476358240
%N A352408 a(n) = [x^(n+1)] (1+x - x^2*C(x^3))^(n*(2*n+1)) / (n*(2*n+1)), for n >= 1, where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108).
%C A352408 This sequence explores yet another property of the Catalan sequence (A000108).
%C A352408 Conjecture: all terms are divisible by 3. Showing that a(n) = 0 (mod 3) for n >= 1 would consequently prove conjectures involving the g.f. G(x) of A352409, namely: G(x) = 1 + x^3*G(x)^2 (mod 3) and G(x) = C(x^3) (mod 3), where C(x) = 1 + x*C(x)^2.
%C A352408 Note: C(x^3) = C(x)^3 (mod 3), thus 1+x - x^2*C(x^3) = 1+x - x^2*C(x)^3 (mod 3) = 1+x + x*(C(x) - C(x)^2) (mod 3) = 2+x - (1-x)*C(x) (mod 3), where C(x) = 1 + x*C(x)^2.
%H A352408 Paul D. Hanna, <a href="/A352408/b352408.txt">Table of n, a(n) for n = 1..300</a>
%F A352408 a(n) ~ 2^(n - 1/2) * exp(n - 1/4) * n^(n - 3/2) / sqrt(Pi). - _Vaclav Kotesovec_, Mar 18 2022
%e A352408 Given C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the Catalan function of A000108, which starts C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + ... + binomial(2*n,n)/(n+1) * x^n + ...
%e A352408 then
%e A352408 1+x - x^2*C(x^3) = 1 + x - x^2 - x^5 - 2*x^8 - 5*x^11 - 14*x^14 - 42*x^17 - 132*x^20 - 429*x^23 - 1430*x^26 - ...
%e A352408 and the table of coefficients of x^k in (1+x - x^2*C(x^3))^(n*(2*n+1)) begin:
%e A352408 n=1: [1, 3, 0, -5, 0, 0, -7, 3, ...];
%e A352408 n=2: [1, 10, 35, 30, -105, -238, 0, 270, ...];
%e A352408 n=3: [1, 21, 189, 910, 2205, 378, -13321, -33324, ...];
%e A352408 n=4: [1, 36, 594, 5880, 38115, 162756, 407862, 175068, ...];
%e A352408 n=5: [1, 55, 1430, 23265, 263835, 2193191, 13612995, 62337330, ...];
%e A352408 n=6: [1, 78, 2925, 70070, 1201200, 15633540, 159772613, 1305975528, ...];
%e A352408 ...
%e A352408 from which the terms a(n) = [x^(n+1)] (1+x - x^2*C(x^3))^(n*(2*n+1))/(n*(2*n+1)) can be derived, for n >= 1, as illustrated by:
%e A352408 a(1) = [x^2] (1+x - x^2*C(x^3))^3 / 3 = 0 / 3 = 0;
%e A352408 a(2) = [x^3] (1+x - x^2*C(x^3))^10 / 10 = 30 / 10 = 3;
%e A352408 a(3) = [x^4] (1+x - x^2*C(x^3))^21 / 21 = 2205 / 21 = 105;
%e A352408 a(4) = [x^5] (1+x - x^2*C(x^3))^36 / 36 = 162756 / 36 = 4521;
%e A352408 a(5) = [x^6] (1+x - x^2*C(x^3))^55 / 55 = 13612995 / 55 = 247509;
%e A352408 a(6) = [x^7] (1+x - x^2*C(x^3))^78 / 78 = 1305975528 / 78 = 16743276;
%e A352408 ...
%e A352408 Apparently, all terms computed this way are divisible by 3 (verified for the initial 1201 terms).
%o A352408 (PARI) {a(n) = my(C3 = (1 - sqrt(1 - 4*x^3 + O(x^(n+3)) ))/(2*x^3));
%o A352408 polcoeff( (1+x - x^2*C3)^(n*(2*n+1)) / (n*(2*n+1)), n+1)}
%o A352408 for(n=1,21, print1(a(n),", " ))
%Y A352408 Cf. A352409.
%K A352408 nonn
%O A352408 1,2
%A A352408 _Paul D. Hanna_, Mar 16 2022